deteriming number of bytes in one line of binary file

조회 수: 13 (최근 30일)
Andrew
Andrew 2014년 6월 9일
댓글: dpb 2014년 6월 9일
Hello,
I am working on translating some old fortran code into MATLAB and have run across the following issue.
In the fortran file I am working with the following check is performed
CHARACTER*72 PICTFILE
CHARACTER*8192 INLINE
INTEGER NPX
INTEGER NLN
INTEGER BYTES
c This is read from another file but I'll just hard code it for now
NPX = 1024
NLN = 1024
bytes=2
open(unit=10, file=pictfile, access='direct', recl=2*npx, status='old')
read(10,rec=nln, err=20) inline(1:2*npx)
go to 21
20 bytes=1
21 continue
close(unit=10)
where nln is the number of lines in the file being read, and npx is the number of integers contained in each line. This check basically determines whether each of those integers is 1 byte or 2 bytes by trying to read in a line of 2 bytes for each integer, and then adjusting to 1 byte if there is an error. I understand the fortran code well enough to figure that out, but now I need to figure out how to perform this check in MATLAB. I have tried using the fgetl command on the file and then reading the length of the characters contained but the length never seems to be more than 4 or 5 characters, when even if each integer is 1 byte the length should be somewhere around 1000.
Can someone help me with devising some way to perform this check in MATLAB? I've thought about maybe a try-catch statement but I'm not very familiar with them and don't really know where to start.
Thanks! Andrew
  댓글 수: 7
Andrew
Andrew 2014년 6월 9일
편집: Andrew 2014년 6월 9일
Hey, from what you asked I input the following fid=fopen('FC21A0004707.DAT') fread(fid,1,'*uint32','b') fclose(fid)
fid=fopen('FC21A0004707.DAT')
fread(fid,1,'*uint32','l')
fclose(fid);
fread(fid,1,'*uint32','l')
and got the following outputs from the file respectively:
ans = 2551174414
ans = 248844184
(I haven't had time to figure out the endianness). This file may not have been made with fortran. It is simply supposed to be image data. I have attached both the file I am trying to read and the fortran function for reading it to this comment. I do have access to gfortran and have been using it to recompile the fortan source I attached with different write statements so that I could check my work.
Also, as for npx, there may be 2*NPX values for each line but only if byte is equal to 2, which is what the above chunk of code checks, then they fold down to simply NPX 16 bit integers. Essentially this is a raw image file that I am trying to read so regardless of whether the integers are 8 bit or 16 bit there will always be exactly 1024 of them.
Also, I apologize that this code is not commented well. I wasn't the one who wrote it I just have to try and interpret it despite never having used fortran before :p...
Andrew
Andrew 2014년 6월 9일
Hey @dpb. I think I figured it out in my answer below. all that the above test is doing is trying to read the last record in the file. If it gets to the end of the file and is still trying to read then it throws an error. However, because of the err=20 option in the read command it doesn't report the error and instead goes to the line identified by 20, thus recording that there can't be 2 bytes per integer because there isn't enough data in the file and then moving on. If the read command doesn't experience an error it just skips over the line from the go to command and keeps bytes set as 2. Certainly no the most robust way to handle this but I supposed that it works.
Thanks for all your help!!
Andrew

댓글을 달려면 로그인하십시오.

채택된 답변

Andrew
Andrew 2014년 6월 9일
편집: Andrew 2014년 6월 9일
For anyone else who may have this problem I believe I figured out a solution using try-catch statements.
I realized that all the above was doing was trying to read the data from the last record in the file. It was set up that if there was an error (it ran out of data in the file) it would kick out and set the number of bytes to 1 instead of 2. In order to do this in Matlab use the following:
fullpath=which(file); %extracting the full file path
s=dir(fullpath); %extracting information about hte file
fid=fopen(file_name,'r'); %opening image file
if s.bytes/NLN==2*NPX %if the file is NLN*NPX*2 bytes
for n=1:NLN %for each line
dn(n,:) = (fread(fid, NPX, '*uint16','b'))'; %reading in lines into DN
end
elseif s.bytes/NLN==NPX %Else if the file is NLN*NPX bytes
for n=1:NLN %for each line
dn(n,:) = (fread(fid, NPX, '*uint8','b'))'; %reading in lines into DN
end
else %If the file is neither something went wrong
error('Invalid file. The file is not the correct size specified by the SUM file')
end
hope this helps, but be careful! This only works if all of the data in the file is the same type (ie uint8 or uint16) and you know the number of entries that should be contained in the file beforehand. Andrew
  댓글 수: 1
dpb
dpb 2014년 6월 9일
... if all of the data in the file is the same type (ie uint8 or uint16) and you know the number of entries that should be contained in the file beforehand.
If that were the case you should be able to just find the SIZE in bytes and divide it is stream ('binary') data only.
It still doesn't make sense unless there was a compiler switch that made the Fortran unformatted record look like a stream file, though, altho apparently it was...does GFortran actually read it correctly w/o any special switch settings? I've not used it enough to know its handling otomh but it's plenty weird if it does.
But, I suppose "all's well that end's well"... :)

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

제품

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by